Question

Two point charges are on the y axis. A 3.10-µC charge is located at y = 1.25 cm, and a -2-µC charge is located at y = −1.80 cm.

(a) Find the total electric potential at the origin.
________ V

(b) Find the total electric potential at the point whose coordinates are (1.50 cm, 0).
________ V

Answer 1

Electric potential is a scalar quantity, which is given by:

V = kQ/R

Part A

Vnet = V1 + V2

Vnet = kQ1/R1 + kQ2/R2

R1 = (1.25 cm - 0) = 0.0125 m

R2 = 0 - (-1.80 cm) = 0.018 m

So,

Vnet = 9*10^9*3.1*10^-6/0.0125 - 9*10^9*2*10^-6/0.018

Vnet = 1.232*10^6 V

Part B

distance between two points is given by

P (x1, y1) and R (x2, y2)

PR = sqrt ((x2 - x1)^2 + (y2 - y1)^2)

Now given points are

for R1, P(0, 1.25), and Q (1.50, 0)

R1 = sqrt (1.50^2 + 1.25^2) = 1.95 cm = 0.0195 m

for R2, P(0, -1.80), and Q (1.50, 0)

R2 = sqrt (1.50^2 + (-1.80)^2) = 2.34 cm = 0.0234 m

Now

Vnet = kQ1/R1 + kQ2/R2

Vnet = 9*10^9*3.1*10^-6/0.0195 - 9*10^9*2*10^-6/0.0234

Vnet = 661538.46 V